A false reasoning for a better intuition behind the universal coefficient theorem
$begingroup$
Let $X$ be a topological space. I am trying to understand the natural homomorphism
$$H^*(M,mathbb{Z}) to H^*(M,mathbb{R}),$$
induced by the inclusion $mathbb{Z} hookrightarrow mathbb{R}$.
The universal coefficient theorem says that this homomorphism may not be injective, namely when torsion elements exist. I am trying to understand this in a very elementary way, and to that aim I have written a false argument of injectivity. I would like to understand where this argument fails, so I could have more intuition behind this non-injectivity. Any help would be appreciated.
Denote $C_k(M)$ the free abelian group generated by $k$-singular chains on $M$, and by:
$$C^k(M,mathbb{Z}) := Hom(C_k(M),mathbb{Z}), quad C^k(M,mathbb{R}) := Hom(C_k(M),mathbb{R}).$$
Then the boundary operator $partial$ on $C_k(M)$ dualizes to boundary operators $delta_{mathbb{Z}}$ and $delta_{mathbb{R}}$ on the two above groups respectively.
Let
$$Z^k_{mathbb{Z}} := ker(delta_{mathbb{Z}} : C^k(M,mathbb{Z}) to C^{k+1}(M,mathbb{Z})), quad Z^k_{mathbb{R}} := ker(delta_{mathbb{R}} : C^k(M,mathbb{R}) to C^{k+1}(M,mathbb{R})),$$
and
$$B^k_{mathbb{Z}} := im(delta_{mathbb{Z}} : C^{k-1}(M,mathbb{Z}) to C^k(M,mathbb{Z})), quad B^k_{mathbb{R}} := im(delta_{mathbb{R}} : C^{k-1}(M,mathbb{R}) to C^k(M,mathbb{R})).$$
- Since $Hom(M,-)$ is left exact, the inclusion $mathbb{Z} hookrightarrow mathbb{R}$ yields an injection $C^k(M,mathbb{Z}) hookrightarrow C^k(M,mathbb{R})$.
- Since $delta_{mathbb{R} | C^k(M,mathbb{Z})} = delta_{mathbb{Z}}$, we have inclusions
$$B^k_{mathbb{Z}} subset B^k_{mathbb{R}}, quad Z^k_{mathbb{Z}} subset Z^k_{mathbb{R}},$$
which yield a natural homomorphism $rho : H^k(M,mathbb{Z}) to H^k(M,mathbb{R})$ in cohomology.
Suppose that $alpha in Z^k_{mathbb{Z}}(M)$ is such that $rho([alpha]) = 0$. This means that $alpha in B^k_{mathbb{R}}$. But $alpha in C^k(M,mathbb{Z})$, so $alpha in B^k_{mathbb{R}} cap C^k(M,mathbb{Z}) = B^k_{mathbb{Z}}$, and therefore $[alpha] = 0$.
Could someone tell me where this reasoning fails ? Thanks a lot !
homology-cohomology
asked Dec 27 '18 at 16:51
BrianTBrianT
1916
$endgroup$
add a comment |
$begingroup$
Let $X$ be a topological space. I am trying to understand the natural homomorphism
$$H^*(M,mathbb{Z}) to H^*(M,mathbb{R}),$$
induced by the inclusion $mathbb{Z} hookrightarrow mathbb{R}$.
The universal coefficient theorem says that this homomorphism may not be injective, namely when torsion elements exist. I am trying to understand this in a very elementary way, and to that aim I have written a false argument of injectivity. I would like to understand where this argument fails, so I could have more intuition behind this non-injectivity. Any help would be appreciated.
Denote $C_k(M)$ the free abelian group generated by $k$-singular chains on $M$, and by:
$$C^k(M,mathbb{Z}) := Hom(C_k(M),mathbb{Z}), quad C^k(M,mathbb{R}) := Hom(C_k(M),mathbb{R}).$$
Then the boundary operator $partial$ on $C_k(M)$ dualizes to boundary operators $delta_{mathbb{Z}}$ and $delta_{mathbb{R}}$ on the two above groups respectively.
Let
$$Z^k_{mathbb{Z}} := ker(delta_{mathbb{Z}} : C^k(M,mathbb{Z}) to C^{k+1}(M,mathbb{Z})), quad Z^k_{mathbb{R}} := ker(delta_{mathbb{R}} : C^k(M,mathbb{R}) to C^{k+1}(M,mathbb{R})),$$
and
$$B^k_{mathbb{Z}} := im(delta_{mathbb{Z}} : C^{k-1}(M,mathbb{Z}) to C^k(M,mathbb{Z})), quad B^k_{mathbb{R}} := im(delta_{mathbb{R}} : C^{k-1}(M,mathbb{R}) to C^k(M,mathbb{R})).$$
- Since $Hom(M,-)$ is left exact, the inclusion $mathbb{Z} hookrightarrow mathbb{R}$ yields an injection $C^k(M,mathbb{Z}) hookrightarrow C^k(M,mathbb{R})$.
- Since $delta_{mathbb{R} | C^k(M,mathbb{Z})} = delta_{mathbb{Z}}$, we have inclusions
$$B^k_{mathbb{Z}} subset B^k_{mathbb{R}}, quad Z^k_{mathbb{Z}} subset Z^k_{mathbb{R}},$$
which yield a natural homomorphism $rho : H^k(M,mathbb{Z}) to H^k(M,mathbb{R})$ in cohomology.
Suppose that $alpha in Z^k_{mathbb{Z}}(M)$ is such that $rho([alpha]) = 0$. This means that $alpha in B^k_{mathbb{R}}$. But $alpha in C^k(M,mathbb{Z})$, so $alpha in B^k_{mathbb{R}} cap C^k(M,mathbb{Z}) = B^k_{mathbb{Z}}$, and therefore $[alpha] = 0$.
Could someone tell me where this reasoning fails ? Thanks a lot !
homology-cohomology
asked Dec 27 '18 at 16:51
BrianTBrianT
1916
$endgroup$
add a comment |
$begingroup$
Let $X$ be a topological space. I am trying to understand the natural homomorphism
$$H^*(M,mathbb{Z}) to H^*(M,mathbb{R}),$$
induced by the inclusion $mathbb{Z} hookrightarrow mathbb{R}$.
The universal coefficient theorem says that this homomorphism may not be injective, namely when torsion elements exist. I am trying to understand this in a very elementary way, and to that aim I have written a false argument of injectivity. I would like to understand where this argument fails, so I could have more intuition behind this non-injectivity. Any help would be appreciated.
Denote $C_k(M)$ the free abelian group generated by $k$-singular chains on $M$, and by:
$$C^k(M,mathbb{Z}) := Hom(C_k(M),mathbb{Z}), quad C^k(M,mathbb{R}) := Hom(C_k(M),mathbb{R}).$$
Then the boundary operator $partial$ on $C_k(M)$ dualizes to boundary operators $delta_{mathbb{Z}}$ and $delta_{mathbb{R}}$ on the two above groups respectively.
Let
$$Z^k_{mathbb{Z}} := ker(delta_{mathbb{Z}} : C^k(M,mathbb{Z}) to C^{k+1}(M,mathbb{Z})), quad Z^k_{mathbb{R}} := ker(delta_{mathbb{R}} : C^k(M,mathbb{R}) to C^{k+1}(M,mathbb{R})),$$
and
$$B^k_{mathbb{Z}} := im(delta_{mathbb{Z}} : C^{k-1}(M,mathbb{Z}) to C^k(M,mathbb{Z})), quad B^k_{mathbb{R}} := im(delta_{mathbb{R}} : C^{k-1}(M,mathbb{R}) to C^k(M,mathbb{R})).$$
- Since $Hom(M,-)$ is left exact, the inclusion $mathbb{Z} hookrightarrow mathbb{R}$ yields an injection $C^k(M,mathbb{Z}) hookrightarrow C^k(M,mathbb{R})$.
- Since $delta_{mathbb{R} | C^k(M,mathbb{Z})} = delta_{mathbb{Z}}$, we have inclusions
$$B^k_{mathbb{Z}} subset B^k_{mathbb{R}}, quad Z^k_{mathbb{Z}} subset Z^k_{mathbb{R}},$$
which yield a natural homomorphism $rho : H^k(M,mathbb{Z}) to H^k(M,mathbb{R})$ in cohomology.
Suppose that $alpha in Z^k_{mathbb{Z}}(M)$ is such that $rho([alpha]) = 0$. This means that $alpha in B^k_{mathbb{R}}$. But $alpha in C^k(M,mathbb{Z})$, so $alpha in B^k_{mathbb{R}} cap C^k(M,mathbb{Z}) = B^k_{mathbb{Z}}$, and therefore $[alpha] = 0$.
Could someone tell me where this reasoning fails ? Thanks a lot !
homology-cohomology
asked Dec 27 '18 at 16:51
BrianTBrianT
1916
$endgroup$
Let $X$ be a topological space. I am trying to understand the natural homomorphism
$$H^*(M,mathbb{Z}) to H^*(M,mathbb{R}),$$
induced by the inclusion $mathbb{Z} hookrightarrow mathbb{R}$.
The universal coefficient theorem says that this homomorphism may not be injective, namely when torsion elements exist. I am trying to understand this in a very elementary way, and to that aim I have written a false argument of injectivity. I would like to understand where this argument fails, so I could have more intuition behind this non-injectivity. Any help would be appreciated.
Denote $C_k(M)$ the free abelian group generated by $k$-singular chains on $M$, and by:
$$C^k(M,mathbb{Z}) := Hom(C_k(M),mathbb{Z}), quad C^k(M,mathbb{R}) := Hom(C_k(M),mathbb{R}).$$
Then the boundary operator $partial$ on $C_k(M)$ dualizes to boundary operators $delta_{mathbb{Z}}$ and $delta_{mathbb{R}}$ on the two above groups respectively.
Let
$$Z^k_{mathbb{Z}} := ker(delta_{mathbb{Z}} : C^k(M,mathbb{Z}) to C^{k+1}(M,mathbb{Z})), quad Z^k_{mathbb{R}} := ker(delta_{mathbb{R}} : C^k(M,mathbb{R}) to C^{k+1}(M,mathbb{R})),$$
and
$$B^k_{mathbb{Z}} := im(delta_{mathbb{Z}} : C^{k-1}(M,mathbb{Z}) to C^k(M,mathbb{Z})), quad B^k_{mathbb{R}} := im(delta_{mathbb{R}} : C^{k-1}(M,mathbb{R}) to C^k(M,mathbb{R})).$$
- Since $Hom(M,-)$ is left exact, the inclusion $mathbb{Z} hookrightarrow mathbb{R}$ yields an injection $C^k(M,mathbb{Z}) hookrightarrow C^k(M,mathbb{R})$.
- Since $delta_{mathbb{R} | C^k(M,mathbb{Z})} = delta_{mathbb{Z}}$, we have inclusions
$$B^k_{mathbb{Z}} subset B^k_{mathbb{R}}, quad Z^k_{mathbb{Z}} subset Z^k_{mathbb{R}},$$
which yield a natural homomorphism $rho : H^k(M,mathbb{Z}) to H^k(M,mathbb{R})$ in cohomology.
Suppose that $alpha in Z^k_{mathbb{Z}}(M)$ is such that $rho([alpha]) = 0$. This means that $alpha in B^k_{mathbb{R}}$. But $alpha in C^k(M,mathbb{Z})$, so $alpha in B^k_{mathbb{R}} cap C^k(M,mathbb{Z}) = B^k_{mathbb{Z}}$, and therefore $[alpha] = 0$.
Could someone tell me where this reasoning fails ? Thanks a lot !
homology-cohomology
homology-cohomology
asked Dec 27 '18 at 16:51
BrianTBrianT
1916
asked Dec 27 '18 at 16:51
BrianTBrianT
1916
asked Dec 27 '18 at 16:51
BrianTBrianT
1916
asked Dec 27 '18 at 16:51
BrianTBrianT
1916
asked Dec 27 '18 at 16:51
BrianTBrianT
1916
1916
add a comment |
add a comment |
1 Answer
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$begingroup$
The error is at the end: the intersection of real boundaries and integral chains is not equal to the integral boundaries.
For instance if you consider a chain complex $mathbb Z overset{2}to mathbb Z$ and change coefficients to $mathbb R$ then in degree 0 degree 1 one has $B^1_{mathbb R} cap C^1_{mathbb Z} = mathbb Z$ but $B^1_{mathbb Z} = 2mathbb Z$.
answered Dec 27 '18 at 17:00
BenBen
4,283617
$endgroup$
$begingroup$
Thanks. Could you extend your comment ? What would be the "difference" between these two sets ?
$endgroup$
– BrianT
Dec 27 '18 at 17:03
$begingroup$
@BrianT I added an example - maybe that helps? If it’s not clear please let me know.
$endgroup$
– Ben
Dec 27 '18 at 17:06
$begingroup$
Thanks a lot. So if I understand well, you identify $mathbb{Z} =C^{-1}(M,mathbb{Z})$ and $mathbb{Z} = C^0(M,mathbb{Z})$ ? Do you have an intuition of why these sets are different ?
$endgroup$
– BrianT
Dec 27 '18 at 17:15
1
$begingroup$
Basically if you rescale a linear map over the integers, it shrinks the image. But if you then pass to the rationals/reals, now division is possible and all rescalings must have the same image. So if you intersect them with the integer values, they all get the same set - ignoring the shrinking image. So, you can’t determine the integer boundaries from the rational/real boundaries alone.
$endgroup$
– Ben
Dec 27 '18 at 17:20
1
$begingroup$
Thanks a lot for your help.
$endgroup$
– BrianT
Dec 27 '18 at 17:22
|
show 6 more comments
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$begingroup$
The error is at the end: the intersection of real boundaries and integral chains is not equal to the integral boundaries.
For instance if you consider a chain complex $mathbb Z overset{2}to mathbb Z$ and change coefficients to $mathbb R$ then in degree 0 degree 1 one has $B^1_{mathbb R} cap C^1_{mathbb Z} = mathbb Z$ but $B^1_{mathbb Z} = 2mathbb Z$.
answered Dec 27 '18 at 17:00
BenBen
4,283617
$endgroup$
$begingroup$
Thanks. Could you extend your comment ? What would be the "difference" between these two sets ?
$endgroup$
– BrianT
Dec 27 '18 at 17:03
$begingroup$
@BrianT I added an example - maybe that helps? If it’s not clear please let me know.
$endgroup$
– Ben
Dec 27 '18 at 17:06
$begingroup$
Thanks a lot. So if I understand well, you identify $mathbb{Z} =C^{-1}(M,mathbb{Z})$ and $mathbb{Z} = C^0(M,mathbb{Z})$ ? Do you have an intuition of why these sets are different ?
$endgroup$
– BrianT
Dec 27 '18 at 17:15
1
$begingroup$
Basically if you rescale a linear map over the integers, it shrinks the image. But if you then pass to the rationals/reals, now division is possible and all rescalings must have the same image. So if you intersect them with the integer values, they all get the same set - ignoring the shrinking image. So, you can’t determine the integer boundaries from the rational/real boundaries alone.
$endgroup$
– Ben
Dec 27 '18 at 17:20
1
$begingroup$
Thanks a lot for your help.
$endgroup$
– BrianT
Dec 27 '18 at 17:22
|
show 6 more comments
$begingroup$
The error is at the end: the intersection of real boundaries and integral chains is not equal to the integral boundaries.
For instance if you consider a chain complex $mathbb Z overset{2}to mathbb Z$ and change coefficients to $mathbb R$ then in degree 0 degree 1 one has $B^1_{mathbb R} cap C^1_{mathbb Z} = mathbb Z$ but $B^1_{mathbb Z} = 2mathbb Z$.
answered Dec 27 '18 at 17:00
BenBen
4,283617
$endgroup$
$begingroup$
Thanks. Could you extend your comment ? What would be the "difference" between these two sets ?
$endgroup$
– BrianT
Dec 27 '18 at 17:03
$begingroup$
@BrianT I added an example - maybe that helps? If it’s not clear please let me know.
$endgroup$
– Ben
Dec 27 '18 at 17:06
$begingroup$
Thanks a lot. So if I understand well, you identify $mathbb{Z} =C^{-1}(M,mathbb{Z})$ and $mathbb{Z} = C^0(M,mathbb{Z})$ ? Do you have an intuition of why these sets are different ?
$endgroup$
– BrianT
Dec 27 '18 at 17:15
1
$begingroup$
Basically if you rescale a linear map over the integers, it shrinks the image. But if you then pass to the rationals/reals, now division is possible and all rescalings must have the same image. So if you intersect them with the integer values, they all get the same set - ignoring the shrinking image. So, you can’t determine the integer boundaries from the rational/real boundaries alone.
$endgroup$
– Ben
Dec 27 '18 at 17:20
1
$begingroup$
Thanks a lot for your help.
$endgroup$
– BrianT
Dec 27 '18 at 17:22
|
show 6 more comments
$begingroup$
The error is at the end: the intersection of real boundaries and integral chains is not equal to the integral boundaries.
For instance if you consider a chain complex $mathbb Z overset{2}to mathbb Z$ and change coefficients to $mathbb R$ then in degree 0 degree 1 one has $B^1_{mathbb R} cap C^1_{mathbb Z} = mathbb Z$ but $B^1_{mathbb Z} = 2mathbb Z$.
answered Dec 27 '18 at 17:00
BenBen
4,283617
$endgroup$
The error is at the end: the intersection of real boundaries and integral chains is not equal to the integral boundaries.
For instance if you consider a chain complex $mathbb Z overset{2}to mathbb Z$ and change coefficients to $mathbb R$ then in degree 0 degree 1 one has $B^1_{mathbb R} cap C^1_{mathbb Z} = mathbb Z$ but $B^1_{mathbb Z} = 2mathbb Z$.
answered Dec 27 '18 at 17:00
BenBen
4,283617
edited Dec 27 '18 at 17:11
answered Dec 27 '18 at 17:00
BenBen
4,283617
answered Dec 27 '18 at 17:00
BenBen
4,283617
answered Dec 27 '18 at 17:00
BenBen
4,283617
4,283617
$begingroup$
Thanks. Could you extend your comment ? What would be the "difference" between these two sets ?
$endgroup$
– BrianT
Dec 27 '18 at 17:03
$begingroup$
@BrianT I added an example - maybe that helps? If it’s not clear please let me know.
$endgroup$
– Ben
Dec 27 '18 at 17:06
$begingroup$
Thanks a lot. So if I understand well, you identify $mathbb{Z} =C^{-1}(M,mathbb{Z})$ and $mathbb{Z} = C^0(M,mathbb{Z})$ ? Do you have an intuition of why these sets are different ?
$endgroup$
– BrianT
Dec 27 '18 at 17:15
1
$begingroup$
Basically if you rescale a linear map over the integers, it shrinks the image. But if you then pass to the rationals/reals, now division is possible and all rescalings must have the same image. So if you intersect them with the integer values, they all get the same set - ignoring the shrinking image. So, you can’t determine the integer boundaries from the rational/real boundaries alone.
$endgroup$
– Ben
Dec 27 '18 at 17:20
1
$begingroup$
Thanks a lot for your help.
$endgroup$
– BrianT
Dec 27 '18 at 17:22
|
show 6 more comments
$begingroup$
Thanks. Could you extend your comment ? What would be the "difference" between these two sets ?
$endgroup$
– BrianT
Dec 27 '18 at 17:03
$begingroup$
@BrianT I added an example - maybe that helps? If it’s not clear please let me know.
$endgroup$
– Ben
Dec 27 '18 at 17:06
$begingroup$
Thanks a lot. So if I understand well, you identify $mathbb{Z} =C^{-1}(M,mathbb{Z})$ and $mathbb{Z} = C^0(M,mathbb{Z})$ ? Do you have an intuition of why these sets are different ?
$endgroup$
– BrianT
Dec 27 '18 at 17:15
1
$begingroup$
Basically if you rescale a linear map over the integers, it shrinks the image. But if you then pass to the rationals/reals, now division is possible and all rescalings must have the same image. So if you intersect them with the integer values, they all get the same set - ignoring the shrinking image. So, you can’t determine the integer boundaries from the rational/real boundaries alone.
$endgroup$
– Ben
Dec 27 '18 at 17:20
1
$begingroup$
Thanks a lot for your help.
$endgroup$
– BrianT
Dec 27 '18 at 17:22
$begingroup$
Thanks. Could you extend your comment ? What would be the "difference" between these two sets ?
$endgroup$
– BrianT
Dec 27 '18 at 17:03
$begingroup$
Thanks. Could you extend your comment ? What would be the "difference" between these two sets ?
$endgroup$
– BrianT
Dec 27 '18 at 17:03
$begingroup$
@BrianT I added an example - maybe that helps? If it’s not clear please let me know.
$endgroup$
– Ben
Dec 27 '18 at 17:06
$begingroup$
@BrianT I added an example - maybe that helps? If it’s not clear please let me know.
$endgroup$
– Ben
Dec 27 '18 at 17:06
$begingroup$
Thanks a lot. So if I understand well, you identify $mathbb{Z} =C^{-1}(M,mathbb{Z})$ and $mathbb{Z} = C^0(M,mathbb{Z})$ ? Do you have an intuition of why these sets are different ?
$endgroup$
– BrianT
Dec 27 '18 at 17:15
$begingroup$
Thanks a lot. So if I understand well, you identify $mathbb{Z} =C^{-1}(M,mathbb{Z})$ and $mathbb{Z} = C^0(M,mathbb{Z})$ ? Do you have an intuition of why these sets are different ?
$endgroup$
– BrianT
Dec 27 '18 at 17:15
1
1
$begingroup$
Basically if you rescale a linear map over the integers, it shrinks the image. But if you then pass to the rationals/reals, now division is possible and all rescalings must have the same image. So if you intersect them with the integer values, they all get the same set - ignoring the shrinking image. So, you can’t determine the integer boundaries from the rational/real boundaries alone.
$endgroup$
– Ben
Dec 27 '18 at 17:20
$begingroup$
Basically if you rescale a linear map over the integers, it shrinks the image. But if you then pass to the rationals/reals, now division is possible and all rescalings must have the same image. So if you intersect them with the integer values, they all get the same set - ignoring the shrinking image. So, you can’t determine the integer boundaries from the rational/real boundaries alone.
$endgroup$
– Ben
Dec 27 '18 at 17:20
1
1
$begingroup$
Thanks a lot for your help.
$endgroup$
– BrianT
Dec 27 '18 at 17:22
$begingroup$
Thanks a lot for your help.
$endgroup$
– BrianT
Dec 27 '18 at 17:22
|
show 6 more comments
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